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## Introduction to systems of equations

Current time:0:00Total duration:2:22

# Systems of equations with graphing: y=7/5x-5 & y=3/5x-1

CCSS Math: 8.EE.C.8, 8.EE.C.8a, HSA.REI.C.6, HSA.REI.D.11

## Video transcript

Just in case we encounter
any more trolls who want us to figure out
what types of money they have in their pockets,
we have devised an exercise for you to practice with. And this is to solve systems
of equations visually. So they say right over here,
graph this system of equations and solve. And they give us two equations. This first one in blue, y
is equal to 7/5x minus 5, and then this one in green,
y is equal to 3/5x minus 1. So let's graph each
of these, and we'll do it in the
corresponding color. So first let's graph
this first equation. So the first thing I see is
its y-intercept is negative 5. Or another way to think about
it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal
to 0, y is going to be equal to negative 5. So that makes sense. And then we see
its slope is 7/5. This was conveniently placed
in slope-intercept form for us. So it's rise over run. So for every time it moves
5 to the right it's, going to move seven up. So if it moves 1, 2,
3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could
have done it is you could have just tested
out some values. You could have said, oh,
when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5
times 5 is 7 minus 5 is 2. So I think we've properly
graphed this top one. Let's try this bottom
one right over here. So we have when x is equal to
0, y is equal to negative 1. So when x is equal to 0,
y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the
right, we will move up 3. So we will go right
over there, and it looks like they intersect
right at that point, right at the point x is
equal to 5, y is equal to 2. So we'll type in x is equal
to 5, y is equal to 2. And you could even verify
by substituting those values into both equations,
to show that it does satisfy both constraints. So let's check our answer. And it worked.